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Behavior of wire subjected by two equal and opposite pulling forces?

  1. May 8, 2012 #1
    Imagine a wire with uniform circular cross-sectional area and length l. It is being pulled at both ends with a horizontal force that is equal and opposite to each other. The wire obeys Hooke's law up to breaking.

    (1) How do we calculate the work done by the wire if both sides of the wire extend by 2e (i.e. an extension of "e" at each end of the wire)? (2) How do I explain the difference between the cancellation of the vector components of each force and the fact that the wire still extends? (3) What is the most proper way of writing down the mathematical working of the extension of the wire? On a side note, (4) How does the energy stored in the wire change, then?

    (5) Also, in what cases are we supposed to assume that the cross-sectional area remains approximately the same and when do we not? (6) Is the approximation e^2 ≈ 0 necessary in this case when I'm calculating the energy stored in the wire?

    Thank you for your help.
  2. jcsd
  3. May 8, 2012 #2


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    You can use [itex] W = \int F.ds[/itex], where F is the tension force and ds is the infinitesimal extension. Since the tension force varies from 0 to T during the extension of 2e, the integral works out to (T/2)(2e) = Te. Or if you don't like integrals, the wire is a spring with force constant k = T/2e (per Hookes law F=kx), so you can use energy methods [itex]W = \Delta PE = 1/2kx^2 = (T/(4e))(4e^2) = Te [/itex], same result.
    The forces must be equal and opposite for equilibrium. Internally, the applied force T, at one end, must be equal to the internal force, T, in the rope, for equilibrium
    extension = FL/AE, where F = the internal tension force and A is the wire cross section and E is the elasticity modulus (note that k=AE/L) , if that is what you mean
    its strain energy increases due to the increase of its PE = 1/2kx^2
    When the wire obeys hookes law in its elastic range, ignore any cross section change, which only becomes significant in the plastic range beyond yield when the wire will 'neck' down
    as long as e is such that the wire still obeys Hooke's law, then the stored energy is valid. Since e is on the order of .001 L, I guess you are right that e^2 must be small
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